1 Answer
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I would start at the bottom. Clearly $1$ divides every member of $L$, so $1$ is the minimum element in the divisibility order on $L$. Now what is the next layer up of the Hasse diagram? It must contain those members of $L$ that are divisible by $1$ (and of course by themselves) but not by any other element of $L$. Those elements are $2,3$, and $5$: every other element of $L$ is divisible by at least one of these three numbers. Thus, your partial Hasse diagram now looks like this:

2 3 5
\ | /
\|/
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What numbers will make up the third row? I claim that the third row contains $4,6,10$, and $15$. First, why not $12,30$, or $60$? $12$ is out because $6$ must be below $12$, and $6$ isn’t in the second row; $30$ is out because $15$ must be below $30$, and $15$ isn’t in the second row; and $60$ is out because every other element of $L$ must be below it. A bit of thought should convince you that no member of $L$ must come below $4,6,10$, or $15$ but above the second row, so that $4,6,10$, and $15$ really do belong in the third row. Specifically, $4$ must lie above $2$; $6$ must lie above $2$ and $3$; $10$ must lie above $2$ and $5$; and $15$ must lie above $3$ and $5$.

You can probably finish it up pretty easily from there: $12$ and $30$ go in the fourth row, with $12$ above $4$ and $3$, and $30$ above $6$ and $5$. Finally, $60$ will be the top element, sitting above $12$ and $30$. As a quick cross-check that $12$ and $30$ are the only members of $L$ that should be immediately below $60$, note that every other member of $L$ (besides $12,30$, and $60$) is a divisor of either $12$ or $30$ and therefore goes somewhere below at least one of these two numbers.